How do you simplify #(sqrt2) + 2 (sqrt2) + (sqrt8) / (sqrt3)#?

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Answer 1 · 2018-07-12T08:10:15

#sqrt(2)/3(9+2sqrt(3))#

#sqrt(2)+2*sqrt(2)+2sqrt(2)/sqrt(3)#
since #sqrt(8)=2sqrt(2)#

#(sqrt(2)`sqrt(3)+2*sqrt(2)*sqrt(3)+2sqrt(2))/sqrt(3) #

multiplying numerator and denominator by #sqrt(3)#

#sqrt(2)/3(3+2*3+2*sqrt(3))#

and this is

#sqrt(2)/3(9+2*sqrt(3))#

Answer 2 · 2018-07-12T08:17:58

#sqrt2+2sqrt2+(sqrt8)/(sqrt3)=color(blue)(3sqrt2+(2sqrt6)/3#

Simplify:

#sqrt2+2sqrt2+(sqrt8)/(sqrt3)#

Simplify #sqrt8#.

#sqrt2+2sqrt2+(sqrt(2xx2xx2))/(sqrt3)#

#sqrt2+2sqrt2+(sqrt(2^2xx2))/(sqrt3)#

Apply square root rule: #sqrt(a^2)=a#

#sqrt2+2sqrt2+(2sqrt2)/(sqrt3)#

Rationalize the denominator by multiplying the numerator and denominator by #sqrt3#.

#sqrt2+2sqrt2+(2sqrt2sqrt3)/(sqrt3sqrt3)#

Apply square root rule: #sqrtasqrta=a#

#sqrt2+2sqrt2+(2sqrt2sqrt3)/3#

Apply square root rule: #sqrtasqrtb=sqrt(ab)#

#sqrt2+2sqrt2+(2sqrt6)/3#

Simplify.

#3sqrt2+(2sqrt6)/3#